The integral is determined using the method of integration by parts :
#color(red)(intu(x)dv(x)=u(x)v(x)-intv(x)du(x))#
Take into consideration the integral of the trigonometric function
#color(red)(intsin(x-1)dx=-cos(x-1))#
Let us compute the given integral:
consider
#color(blue)(u(x)=x^2rArrd(u(x))=2xdx)#
then,
#color(brown)(dv(x)=sin(x-1)rArrv(x)=-cos(x-1))#
#intx^2sin(x-1)dx# integral 1
#=color(red)(u(x)*v(x)-intdu(x)v(x))#
#=color(blue)x^2*color(brown)(-cos(x-1))-intcolor(blue)(2xdx)color(brown)((-cos(x-1))#
#=-x^2cos(x-1)+2color(purple)(intxcos(x-1)dx)#
#color(purple)(intxcos(x-1)dx)#
let #a(x)=xrArrda(x)=dx#
then #db(x)=cos(x-1)rArrb(x)=sin(x-1)#
#color(purple)(intxcos(x-1)dx)#
#=color(red)(a(x)*b(x)-intb(x)da(x))#
#=xsin(x-1)-intsin(x-1)dx#
#=xsin(x-1)-intd(-cos(x-1))#
#=xsin(x-1)+intdcos(x-1)#
#color(purple)(=xsin(x-1)+cos(x-1))#
Let us substitute #color(purple)(intxcos(x-1)dx)#
in integral 1
#intx^2sin(x-1)dx#
#=-x^2cos(x-1)+2color(purple)(intxcos(x-1)dx)#
#=-x^2cos(x-1)+2(xsin(x-1)+cos(x-1))#
#=-x^2cos(x-1)+2xsin(x-1)+2cos(x-1)#
Therefore,
#intx^2sin(x-1)dx#
#=-x^2cos(x-1)+2xsin(x-1)+2cos(x-1)#
